A particle executes simple harmonic motion such that the graph of its velocity v against its displacement x is an ellipse. If the magnitude of the intercept of this ellipse on the velocity axis is v₀ and on the displacement axis is x₀ , what is the maximum acceleration of the particle?

Step-by-step solution

The velocity v and displacement x of a particle in simple harmonic motion are related by the equation of an ellipse: v^2 ( A)^2 + x^2 A^2 = 1 The intercept on the velocity axis corresponds to the maximum velocity, so v₀ = A . The intercept on the displacement axis corresponds to the amplitude, so x₀ = A . From these, the angular frequency is: = v₀ x₀ The maximum acceleration of a particle in simple harmonic motion is given by a_ max = ^2 A . Substituting the expressions for and A : a_ max = ( v₀ x₀ )^2 x₀ = v₀^2 x₀